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If n = 4p, where p is a prime number greater than 2, how man

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If n = 4p, where p is a prime number greater than 2, how man [#permalink] New post 27 Dec 2012, 04:55
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If n = 4p, where p is a prime number greater than 2, how many different positive even divisors does n have, including n ?

(A) Two
(B) Three
(C) Four
(D) Six
(E) Eight
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Re: If n = 4p, where p is a prime number greater than 2, how man [#permalink] New post 27 Dec 2012, 04:59
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Walkabout wrote:
If n = 4p, where p is a prime number greater than 2, how many different positive even divisors does n have, including n ?

(A) Two
(B) Three
(C) Four
(D) Six
(E) Eight


Since we cannot have two correct answers just pick a prime greater than 2, and see how many different positive even divisors will 4p have.

p = 3 --> 4p = 12--> 12 has 4 even divisors: 2, 4, 6, and 12.

Answer: C.

Or this way: since p is prime greater than 2, then p=odd, thus 4p=even, which means that it has 4 even divisors: 2, 4, 2p, and 4p.

Similar question to practice: if-n-is-a-prime-number-greater-than-3-what-is-the-remainder-137869.html

Hope it helps.
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Re: If n = 4p, where p is a prime number greater than 2, how man [#permalink] New post 12 Apr 2013, 23:08
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Bunuel wrote:
Walkabout wrote:
If n = 4p, where p is a prime number greater than 2, how many different positive even divisors does n have, including n ?

(A) Two
(B) Three
(C) Four
(D) Six
(E) Eight


Since we cannot have two correct answers just pick a prime greater than 2, and see how many different positive even divisors will 4p have.

p = 3 --> 4p = 12--> 12 has 4 even divisors: 2, 4, 6, and 12.

Answer: C.

Or this way: since p is prime greater than 2, then p=odd, thus 4p=odd, which means that it has 4 even divisors: 2, 4, 2p, and 4p.

Similar question to practice: if-n-is-a-prime-number-greater-than-3-what-is-the-remainder-137869.html

Hope it helps.


Hi Bunnel
Just a small typo in the explanation
Or this way: since p is prime greater than 2, then p=odd, thus 4p=odd, which means that it has 4 even divisors: 2, 4, 2p, and 4p.
It should be 4p = Even
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Re: If n = 4p, where p is a prime number greater than 2, how man [#permalink] New post 13 Apr 2013, 02:49
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Dipankar6435 wrote:
Bunuel wrote:
Walkabout wrote:
If n = 4p, where p is a prime number greater than 2, how many different positive even divisors does n have, including n ?

(A) Two
(B) Three
(C) Four
(D) Six
(E) Eight


Since we cannot have two correct answers just pick a prime greater than 2, and see how many different positive even divisors will 4p have.

p = 3 --> 4p = 12--> 12 has 4 even divisors: 2, 4, 6, and 12.

Answer: C.

Or this way: since p is prime greater than 2, then p=odd, thus 4p=odd, which means that it has 4 even divisors: 2, 4, 2p, and 4p.

Similar question to practice: if-n-is-a-prime-number-greater-than-3-what-is-the-remainder-137869.html

Hope it helps.


Hi Bunnel
Just a small typo in the explanation
Or this way: since p is prime greater than 2, then p=odd, thus 4p=odd, which means that it has 4 even divisors: 2, 4, 2p, and 4p.
It should be 4p = Even
Cheers


Typo edited. Thank you. +1.
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Re: If n = 4p, where p is a prime number greater than 2, how man [#permalink] New post 20 Apr 2013, 23:04
Is one not a divisor?
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Re: If n = 4p, where p is a prime number greater than 2, how man [#permalink] New post 21 Apr 2013, 03:06
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Re: If n = 4p, where p is a prime number greater than 2, how man [#permalink] New post 04 Mar 2014, 02:45
p can get any prime number greater than 2 so there should be unlimited different divisor for the n because p can be 11,13,17,19....

What is wrong with this ?
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Re: If n = 4p, where p is a prime number greater than 2, how man [#permalink] New post 04 Mar 2014, 02:53
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lool wrote:
If n = 4p, where p is a prime number greater than 2, how many different positive even divisors does n have, including n ?

(A) Two
(B) Three
(C) Four
(D) Six
(E) Eight

p can get any prime number greater than 2 so there should be unlimited different divisor for the n because p can be 11,13,17,19....

What is wrong with this ?


No matter which prime p is, 4p will have only four EVEN divisors: 2, 4, 2p, and 4p. Try to check it with any prime greater than 2.

Does this make sense?
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Re: If n = 4p, where p is a prime number greater than 2, how man [#permalink] New post 04 Mar 2014, 20:19
lool wrote:
p can get any prime number greater than 2 so there should be unlimited different divisor for the n because p can be 11,13,17,19....

What is wrong with this ?



4 has two even divisors >> 2 & 4

Any prime no p divisors will have 2p & 4p (from above)

So total = 4
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Re: If n = 4p, where p is a prime number greater than 2, how man [#permalink] New post 19 Apr 2014, 22:35
n=4*p

And p(prime) > 2

n= 2*2*p

Positive even divisors 'n' can have including 'n':

2
2p
4
4p which is also equal to 'n'

Hence 4 positive even divisors.
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Re: If n = 4p, where p is a prime number greater than 2, how man [#permalink] New post 01 Jun 2014, 21:06
n= 4p = 2^2*p^1
so total no. of factors: (2+1)*(1+1)= 6
total no. of odd factors, since p is odd as it is a prime>2: p^1 and p^0 = 2
Total no. of even factors: 6 - 2 = 4

Now if n was n=4pq where p and q are both prime no.s greater than 2 then:
total no. of factors: (2+1)*(1+1)*(1+1)= 12
total no. of odd factors, since p is odd as it is a prime>2: (1+1)*(1+1)= 4
Total no. of even factors: 12 - 4 = 8

Hi Bunuel, could you validate my logic pls?
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Last edited by MensaNumber on 02 Jun 2014, 00:59, edited 1 time in total.
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Re: If n = 4p, where p is a prime number greater than 2, how man [#permalink] New post 01 Jun 2014, 23:04
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MensaNumber wrote:
n= 4p = 2^2*p^1
so total no. of factors: (2+1)*(1+1)= 6
total no. of odd factors, since p is odd as it is a prime>2: p^1 and p^0 = 2
Total no. of even factors: 6 - 2 = 4

Now if n was n=4pq where p and q are both prime no.s greater than 2 then:
total no. of factors: (2+1)*(1+1)*(1+1)= 12
total no. of odd factors, since p is odd as it is a prime>2: (1+1)*(1+1)= 4
Total no. of even factors: 12 - 4 = 8

Hi Bunuel, could you validate my logic pls?


Yes, your approach is correct.
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Re: If n = 4p, where p is a prime number greater than 2, how man   [#permalink] 01 Jun 2014, 23:04
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